JEE Physics 2025 Question with Solution

Given: mass of wire $m = (0.60 \pm 0.003) \, \text{g}$, radius $r = (0.50 \pm 0.01) \, \text{cm}$, length $L = (10.00 \pm 0.05) \, \text{cm}$.

Find: maximum percentage error in density $\rho$.

For a wire,

$$\rho = \frac{m}{\pi r^2 L}$$

So the maximum fractional error is

$$\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta L}{L}$$

Now calculate each percentage error:

$$\frac{\Delta m}{m} \times 100 = \frac{0.003}{0.60} \times 100 = 0.5\%$$ $$\frac{\Delta r}{r} \times 100 = \frac{0.01}{0.50} \times 100 = 2\%$$ $$\frac{\Delta L}{L} \times 100 = \frac{0.05}{10.00} \times 100 = 0.5\%$$

Therefore, total percentage error in density is

$$0.5\% + 2(2\%) + 0.5\% = 5\%$$

Therefore, the maximum percentage error in the density is $5\%$. The correct option is B.

Given: $\rho = \frac{m}{\pi r^2 l}$ with measured values $m = 0.60 \, \text{g}$, $r = 0.50 \, \text{cm}$, and $l = 10.00 \, \text{cm}$ along with absolute errors $0.003 \, \text{g}$, $0.01 \, \text{cm}$, and $0.05 \, \text{cm}$ respectively.

Find: maximum percentage error in $\rho$.

Since density depends directly on $m$ and inversely on $r^2$ and $l$,

$$\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}$$

Substitute the given values:

$$\frac{\Delta m}{m} = \frac{0.003}{0.60} = 0.005 = 0.5\%$$ $$\frac{\Delta r}{r} = \frac{0.01}{0.50} = 0.02 = 2\%$$ $$\frac{\Delta l}{l} = \frac{0.05}{10.00} = 0.005 = 0.5\%$$

Hence,

$$\text{Total percentage error} = 0.5 + 2(2) + 0.5 = 5\%$$

Thus, the maximum percentage error in density is $5\%$.